If the function y = a log x + bx^2 + x has ex | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $y = a \log x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values at $x = 1$ an

Vedclass · Accepted Answer

$\left( -\frac{2}{3}, -\frac{1}{6} \right)$

Vedclass · Answer

(D) Given $y = a \log x + bx^2 + x$. Find the derivative: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extreme values at $x = 1$ and $x = 2$,the derivative must be zero at these points. At $x = 1$: $\frac{a}{1} + 2b(1) + 1 = 0 \implies a + 2b = -1$ (Equation $1$). At $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b = -1 \implies a + 8b = -2$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(a + 8b) - (a + 2b) = -2 - (-1) \implies 6b = -1 \implies b = -\frac{1}{6}$. Substitute $b = -\frac{1}{6}$ into Equation $1$: $a + 2(-\frac{1}{6}) = -1 \implies a - \frac{1}{3} = -1 \implies a = -1 + \frac{1}{3} = -\frac{2}{3}$. Thus,$(a, b) = \left( -\frac{2}{3}, -\frac{1}{6} \right)$.

If the function $y = a \log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$,then $(a, b) = \dots$

Similar Questions

If the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$,where $a > 0$,attains its maximum and minimum at $p$ and $q$ respectively such that $p^2 = q$,then $a$ equals:

Consider a cuboid of sides $2x$,$4x$,and $5x$ and a closed hemisphere of radius $r$. If the sum of their surface areas is a constant $k$,then the ratio $x:r$,for which the sum of their volumes is maximum,is

The maximum value of $[x(x-1)+1]^{\frac{1}{3}}$ for $0 \leq x \leq 1$ is

The maximum value $M$ of $f(x) = 3^x + 5^x - 9^x + 15^x - 25^x$,as $x$ varies over all real numbers,satisfies:

The point on the curve $x^2 = 2y$ which is nearest to the point $(0, 5)$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module